Страница 262 номер 1039, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

\(\begin{cases} b_{1} + b_{4} = \frac{35}{3} \\ b_{2} + b_{3} = 10 \end{cases}\ \Rightarrow \begin{cases} b_{1} + b_{1}q^{3} = \frac{35}{3} \\ b_{1}q + b_{1}q^{2} = 10 \end{cases}\ \Rightarrow \begin{cases} b_{1}\left( 1 + q^{3} \right) = \frac{35}{3} \\ b_{1}q(1 + q) = 10 \end{cases}\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & b_{1}(1 + q)\left( 1 - q + q^{2} \right) = \dfrac{35}{3} \\ & b_{1}q(1 + q) = 10 \end{aligned} \right.\ \Rightarrow \begin{cases} \frac{10}{q}\left( 1 - q + q^{2} \right) = \frac{35}{3} \\ b_{1}(1 + q) = \frac{10}{q} \end{cases}\ \Rightarrow\)
\(\dfrac{10}{q}\left( 1 - q + q^{2} \right) = \dfrac{35}{3}\)
\(30\left( 1 - q + q^{2} \right) = 35q\)
\(6 - 6q + 6q^{2} = 7q\)
\(6q^{2} - 13q + 6 = 0\)
\(D = 169 - 4 \cdot 36 = 169 - 144 = 25\)
\(q = \dfrac{13 - 5}{12} = \dfrac{8}{12} = \dfrac{2}{3} \Rightarrow b_{1} = \dfrac{10}{q(1 + q)} = \dfrac{10}{\dfrac{2}{3}\left( 1 + \dfrac{2}{3} \right)} = 10 \cdot \dfrac{3}{2} \cdot \dfrac{3}{5} = 9\)
\(b_{2} = b_{1}q = 9 \cdot \dfrac{2}{3} = 6\)
\(b_{3} = b_{2}q = 6 \cdot \dfrac{2}{3} = 4\)
\(b_{4} = b_{3}q = 4 \cdot \dfrac{2}{3} = \dfrac{8}{3}\)
\(q = \dfrac{13 + 5}{12} = \dfrac{18}{12} = \dfrac{3}{2} \Rightarrow b_{1} = \dfrac{10}{q(1 + q)} = \dfrac{10}{\dfrac{3}{2} \cdot \left( 1 + \dfrac{3}{2} \right)} = 10 \cdot \dfrac{2}{3} \cdot \dfrac{2}{5} = \dfrac{8}{3}\)
\(b_{2} = b_{1}q = \dfrac{8}{3} \cdot \dfrac{3}{2} = 4\)
\(b_{3} = b_{2}q = 4 \cdot \dfrac{3}{2} = 6\)
\(b_{4} = b_{3}q = 6 \cdot \dfrac{3}{2} = 9\)

Ответ: \(\frac{8}{3};\ 4;\ 6;\ 9\).